How do you differentiate #f(t)=1/2t^6-3t^4+1#?

1 Answer
Oct 30, 2016

We can differentiate term by term. (Things that are added together are called "terms".)

So we need to differentiate #1/2t^6# and #-3t^4# and #1#. (We don't need to write all of this, but I'm explaining our thought process.)

To differentiate #1/2 t^6# we will used the power rule. The constant #1/2# just hangs out out front.
In more proper language: the derivative of #1/2# times #t^6# is #1/2# times the derivative of #t^6#.

In notation: #d/dt(1/2t^6) = 1/2 d/dt(t^6)#

Now, we find the derivative of #t^^6#. Multiply by the exponent, then subtract one from the exponent to get the new exponent.

#6t^(6-1) = 6t^5#

We use the same process to differentiate #-3t^4#.

The derivative of a constant, like #1#, is #0#

The derivative of #f(t) = 1/2t^6-3t^4+1# is

#f'(t) = 1/2(6t^5) - 3(4t^3) +0# #" "# (we often skip writing this)

#f'(t) = 3t^5-12t^3#